The following equilibria were obtained at 609 K:

Question

The following equilibria were obtained at 609 K:

CoO(s) + H₂(g) ↔ Co(s) + H₂O(g)   K_c = 77

CoO(s) + CO(g) ↔ Co(s) + CO₂(g)   K_c = 227

Based on these equilibria, calculate the equilibrium constant for:

CoO(s) + 2H₂(g) + CO₂(g) ↔ Co(s) + 2H₂O(g) + CO(g)

Answer 1

CoO(s) + 2H₂(g) + CO₂(g) <==> Co(s) + 2H₂O(g) + CO(g) TARGET EQUATION

1). CoO(s) + H₂(g) <=> Co(s) + H₂O(g)   K_c = 77

2). CoO(s) + CO(g) <=> Co(s) + CO₂(g)   K_c = 227

multiply (1) by 2: 2CoO + 2H₂ ==> 2Co + 2H₂O Kc = (77)² = 5929 (square K when doubling amts)

reverse (2) to get: Co + CO₂ ==> CoO + CO Kc = 1/227 = 0.00441 (take reciprocal of K when reverse)

Add (1) and (2) together...

2CoO + 2H₂ + Co + CO₂ ==> 2Co + 2H₂O + CoO + CO and cancel common items to obtain...

CoO + 2H₂ + CO₂ ==> Co + 2H₂O + CO TARGET EQUATION

To get the K for the added equation, we must multiply the 2 K values from above...

5929 x 0.00441 = 26.1 = K (when adding 2 equations, you multiply the K values)